Squarefree Factorisations

Let's n an integer with the following factorisation : $n=a_1^{e_1}\times a_2^{e_2}\times \cdots \times a_p^{e_p}$ where $a_i$ are squarefree and $\mathrm{\forall }i\in \{1,\dots ,p-1\}{a}_i\text{ divides }{a}_{i+1}$

For instance :
$56=2^2\times {14}^1$
$5040=2^2\times 6^1\times {210}^1$
$526773121875=3^2\times {15}^3\times {1155}^1\times {15015}^1$

It can be proved that this factorisation is unique.

For such a factorisation, let's consider all the divisors of n : $a_1^{f_1}\times a_2^{f_2}\times \dots a_p^{f_p}\text{ where }0\le f_i\le e_i$

Define ${\sigma }^{\prime }(n)=\sum _d(d)$ where d runs over the divisors of n as defined above
${\sigma }^{\prime }(5040)=1+2+4+6+12+24+210+420+840+1260+2520+5040=10339$

We say that n is a champion if the ratio $\frac{{\sigma }^{\prime }(n)}{n}$ is greater than any ratio $\frac{{\sigma }^{\prime }(m)}{m}$ with $m<n$

Here are the first 10 champions:
$1\cdot 1\Rightarrow 2$
$2\cdot 24\Rightarrow 2,04166666666667$
$3\cdot 48\Rightarrow 2,1875$
$4\cdot 96\Rightarrow 2,26041666666667$
$5\cdot 192\Rightarrow 2,296875$
$6\cdot 384\Rightarrow 2,3151041$6666667
$7\cdot 768\Rightarrow 2,32421875$
$8\cdot 1152\Rightarrow 2,3515625$
$9\cdot 2304\Rightarrow 2,37022569444444$
$10\cdot 4608\Rightarrow 2,37955729166667$

What is the ${66}^{th}$ champion?

[My timing: 5 sec]